\(\int \frac {(a^2+2 a b x^3+b^2 x^6)^p}{x^5} \, dx\) [137]

Optimal result

Integrand size = 24, antiderivative size = 60 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^p}{x^5} \, dx=-\frac {\left (1+\frac {b x^3}{a}\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},-2 p,-\frac {1}{3},-\frac {b x^3}{a}\right )}{4 x^4} \]

[Out]

-1/4*(b^2*x^6+2*a*b*x^3+a^2)^p*hypergeom([-4/3, -2*p],[-1/3],-b*x^3/a)/x^4/((1+b*x^3/a)^(2*p))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1370, 371} \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^p}{x^5} \, dx=-\frac {\left (\frac {b x^3}{a}+1\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},-2 p,-\frac {1}{3},-\frac {b x^3}{a}\right )}{4 x^4} \]

[In]

Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^p/x^5,x]

[Out]

-1/4*((a^2 + 2*a*b*x^3 + b^2*x^6)^p*Hypergeometric2F1[-4/3, -2*p, -1/3, -((b*x^3)/a)])/(x^4*(1 + (b*x^3)/a)^(2
*p))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1370

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a
+ b*x^n + c*x^(2*n))^FracPart[p]/(1 + 2*c*(x^n/b))^(2*FracPart[p])), Int[(d*x)^m*(1 + 2*c*(x^n/b))^(2*p), x],
x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = \left (\left (1+\frac {b x^3}{a}\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p\right ) \int \frac {\left (1+\frac {b x^3}{a}\right )^{2 p}}{x^5} \, dx \\ & = -\frac {\left (1+\frac {b x^3}{a}\right )^{-2 p} \left (a^2+2 a b x^3+b^2 x^6\right )^p \, _2F_1\left (-\frac {4}{3},-2 p;-\frac {1}{3};-\frac {b x^3}{a}\right )}{4 x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^p}{x^5} \, dx=-\frac {\left (\left (a+b x^3\right )^2\right )^p \left (1+\frac {b x^3}{a}\right )^{-2 p} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},-2 p,-\frac {1}{3},-\frac {b x^3}{a}\right )}{4 x^4} \]

[In]

Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^p/x^5,x]

[Out]

-1/4*(((a + b*x^3)^2)^p*Hypergeometric2F1[-4/3, -2*p, -1/3, -((b*x^3)/a)])/(x^4*(1 + (b*x^3)/a)^(2*p))

Maple [F]

\[\int \frac {\left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )^{p}}{x^{5}}d x\]

[In]

int((b^2*x^6+2*a*b*x^3+a^2)^p/x^5,x)

[Out]

int((b^2*x^6+2*a*b*x^3+a^2)^p/x^5,x)

Fricas [F]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^p}{x^5} \, dx=\int { \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{x^{5}} \,d x } \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^p/x^5,x, algorithm="fricas")

[Out]

integral((b^2*x^6 + 2*a*b*x^3 + a^2)^p/x^5, x)

Sympy [F]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^p}{x^5} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{p}}{x^{5}}\, dx \]

[In]

integrate((b**2*x**6+2*a*b*x**3+a**2)**p/x**5,x)

[Out]

Integral(((a + b*x**3)**2)**p/x**5, x)

Maxima [F]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^p}{x^5} \, dx=\int { \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{x^{5}} \,d x } \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^p/x^5,x, algorithm="maxima")

[Out]

integrate((b^2*x^6 + 2*a*b*x^3 + a^2)^p/x^5, x)

Giac [F]

\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^p}{x^5} \, dx=\int { \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{p}}{x^{5}} \,d x } \]

[In]

integrate((b^2*x^6+2*a*b*x^3+a^2)^p/x^5,x, algorithm="giac")

[Out]

integrate((b^2*x^6 + 2*a*b*x^3 + a^2)^p/x^5, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^p}{x^5} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^p}{x^5} \,d x \]

[In]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^p/x^5,x)

[Out]

int((a^2 + b^2*x^6 + 2*a*b*x^3)^p/x^5, x)